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Maths Question 17 – JEE-MAIN 2026

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The value of ∫020π(sin4⁡x+cos4⁡x)dx is equal to :

Begin by simplifying the expression sin4⁡x+cos4⁡x using trigonometric identities to make the integration more manageable.

● Video Walkthrough
Step 1: Simplify the Integrand✦ Active

Rewrite the expression sin4⁡x+cos4⁡x using algebraic and trigonometric identities.

sin4⁡x+cos4⁡x=(sin2⁡x+cos2⁡x)2−2sin2⁡xcos2⁡x=1−2(sin⁡xcos⁡x)2=1−2(sin⁡(2x)2)2=1−12sin2⁡(2x)Usingsin2⁡θ=1−cos⁡(2θ)2,weget:=1−12(1−cos⁡(4x)2)=1−14+14cos⁡(4x)=34+14cos⁡(4x)
Step 2: Integrate the Simplified Expression○ Expand

Substitute the simplified integrand back into the definite integral.

I=∫020π(34+14cos⁡(4x))dx=∫020π34dx+14∫020πcos⁡(4x)dx
Step 3: Evaluate the Definite Integral○ Expand

Calculate each part of the integral. The integral of cos⁡(4x) over 20π (which is a multiple of its period 2π4=π2) is zero.

∫020π34dx=[34x]020π=34(20π−0)=15π∫020πcos⁡(4x)dx=0Therefore,I=15π+14(0)=15π
💡 Teacher's Secret Hint

Remember that the integral of a cosine function over an interval that is an integer multiple of its period is zero.

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